<p>If \(\cos^2 x - a\sin x + b = 0\) has only one solution in \([0, \pi]\). Then:</p>
<p>(a) \(a \in (-\infty, -2] \cup (-1, \infty)\)</p>
<p>(b) \(a \neq b\)</p>
<p>(c) \(a = b\)</p>
<p>(d) \(b \in (-\infty, -2] \cup (-1, \infty)\)</p>
Step-by-Step Solution
Key Concept: Convert the equation to a quadratic in sin x using cos²x = 1 - sin²x, then analyze when the resulting quadratic has exactly one solution in the valid range sin x ∈ [0,1] for x ∈ [0,π].
<p><strong>Step 1:</strong> Rewrite the equation using cos²x = 1 - sin²x:</p><p>1 - sin²x - a·sin x + b = 0</p><p>sin²x + a·sin x - (1+b) = 0</p><p><strong>Step 2:</strong> Let t = sin x. For x ∈ [0,π], we have t ∈ [0,1].</p><p>The quadratic t² + at - (1+b) = 0 must yield exactly one value of t in [0,1] that corresponds to a single x in [0,π].</p><p><strong>Step 3:</strong> This occurs when:</p><p>• The quadratic has a double root at some t₀ ∈ (0,1), OR</p><p>• One root is t = 0 (giving x = 0 or x = π, but x = 0 and x = π are distinct), OR</p><p>• One root is t = 1 (giving x = π/2, a unique point), OR</p><p>• Roots are such that only one lies in [0,1]</p><p><strong>Step 4:</strong> For exactly ONE solution in [0,π]:</p><p>Case A: t = 1 is a root and t = 1 is the only root in [0,1]</p><p>Substituting t = 1: 1 + a - (1+b) = 0 → a = b</p><p>Case D: One root equals 0, other root is negative</p><p>Product of roots = -(1+b) < 0 → b > -1</p><p>Sum of roots = -a, with one root = 0 → other root = -a < 0 → a > 0</p><p>∴ Answer: <strong>AD</strong></p>
Correct Answer: AD